Question

Design a wideband band-pass active filter that has a gain of 17dB in the passbandand has corner frequencies at the lower and higher frequency limits of human hearing (i.e. 20Hz to 20kHz).

Answer 1

Solution :

A wide band pass filter has (Q < 10)

Gain in dB of overall filter = 17 dB

$F_L = 20 \ Hz$

$F_H = 20 \ kHz$

$17 = 10 \log A_0$

$1.7 = \log A_0$

$A_0 = 5.47$

$F_L = 20 \ Hz$

$F_L = \frac{1}{2 \pi R_1C_1}$

Let $C_1 = 0.1 \mu F$

$20 = \frac{1}{2 \pi \times R_1 \times 0.1 \times 10^{-6}}$

$R_1 = \frac{10^6}{4 \pi \times 0.1}$

$R_1 = 79.6 \ k \Omega$

$F_H = 20 \ kHz$

$F_H = \frac{1}{2 \pi R_2C_2}$

$20 \times 10^3 = \frac{1}{2 \pi R_2 \times 0.1 \times 10^{-6}}$

$R_2 = \frac{10^3}{4 \pi \times 10^3}$

$R_2 = 79.6 \ \Omega$

Since the overall gain off filter is 5.47

$A_0 = 5.47$

$A_0 = A_{01} \times A_{02}$

$A_0 = 3.4 \times 1.6$

$A_0 \sim 5.47$

$A_{01} = 1+ \frac{R_F_1}{R_i_1}$

$3.4 = 1+ \frac{R_F_1}{R_i_1}$

$2.4 = \frac{10 k}{R_i_1}$

$R_i_1 = 4166 \ \Omega$

$R_i_1 = 4.1 \ k \Omega$

$A_{02} = 1+ \frac{R_F_2}{R_i_2}$

$1.6 = 1+ \frac{R_F_2}{R_i_2}$

$0.6 = \frac{R_F_2}{R_i_1}$

$R_i_2 = \frac{0.1 k}{0.6}$

$R_i_2 = 166 \ \Omega$